How many grams of gas are present in a sample that has a molar mass of 70.0 g/mol and occupies a 2.00 l container at 117 kpa and 35.1 degrees celsius?

The ideal gas law of gases states that the volume occupied by certain moles of a gas has a pressure at a temperature in kelvin. The law combines both boyle's law and the charle's law. Therefore, PV=nRT where P is the pressure, V is the volume, n is the number of moles, T is the temperature and R is the gas constant. 1 atm = 101.325 kPa, therefore; 117kPa = 1.155 atm, 35.1 °c = 308.25 Kelvin and R = 0.0821

PV=nRT

n = PV/RT

= (1.155 * 2) / (0.0821 * 308.25)

= 0.0913 moles

But, 1 mole = 70 g

Therefore, 0.0913 moles * 70 = 6.39 g